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K2HPO4 is a highly water-soluble salt. It can be used as a buffering agent, food additive, or
fertilizer. The neutralization of phosphoric acid with potassium hydroxide can be used to form
potassium hydrogen phosphate.
H3PO4 + 2KOH K2HPO4 + 2H2O
100 g of phosphoric acid (H3PO4) is reacted with excess potassium hydroxide. The final mass
of K2HPO4 produced is found to be 163 g.
What is the percent yield?

Respuesta :

Answer:

91.7%

Explanation:

The percent yield when it could be used as a buffering agent should be considered as the 91.74%.

Calculation of the percent yield:

Since we know that

Number of moles = Mass in gram / Molar mass

And,

Molar mass of H3PO4 = 97.994 g/mol

So,

Number of moles of H3PO4

= 100 g / 97.994 g/mol

= 1.02 mol

Now

Molar mass of K2HPO4 = 174.2 g/mol

So,

Theoretical yield of K2HPO4 = 174.2 g/mol * 1.02 mol

= 177.684 g

So finally

Percent yield = Actual yield * 100 / Theoretical yield

= 163 g * 100 / 177.684 g

= 91.74%

hence, The percent yield when it could be used as a buffering agent should be considered as the 91.74%.

learn more about percentage here: https://brainly.com/question/14849541

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